How do you solve Frequency Tracker on LeetCode?

Frequency Tracker (LeetCode #2671) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(n) space complexity. Key insight: Using hash maps allows for efficient frequency tracking.

What is the brute force solution for Frequency Tracker?

The brute force approach for Frequency Tracker is Brute Force, with O(n²) time complexity and O(n) space complexity. In the brute force approach, we maintain a list of numbers and their frequencies by iterating through the list every time we need to add or delete a number. This is straightforward but inefficient as the size of the list grows. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Frequency Tracker?

Frequency Tracker uses the following concepts: Hash Table, Design. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Frequency Tracker?

Explain your thought process clearly while coding. Consider edge cases like deleting a number that doesn't exist. Discuss the trade-offs of different data structures you could use.

What are common mistakes when solving Frequency Tracker?

Not updating both frequency maps correctly after add/delete operations. Assuming that a number will always exist when trying to delete.

#2671

Frequency Tracker

Medium

Hash TableDesignHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(n)	O(n)

💡

Intuition

Time O(1)Space O(n)

The optimal solution uses two hash maps: one to track the frequency of each number and another to track how many numbers have a specific frequency. This allows for efficient updates and queries.

⚙️

Algorithm

5 steps

1Step 1: Use a hash map to store the frequency of each number.
2Step 2: Use another hash map to store how many numbers have each frequency.
3Step 3: For add, update the frequency of the number and adjust the frequency counts accordingly.
4Step 4: For deleteOne, decrease the frequency and adjust the frequency counts.
5Step 5: For hasFrequency, simply check if the frequency exists in the frequency map.

solution.py29 lines

1class FrequencyTracker:
2    def __init__(self):
3        self.freq_map = {}
4        self.count_map = {}
5
6    def add(self, number):
7        old_freq = self.freq_map.get(number, 0)
8        new_freq = old_freq + 1
9        self.freq_map[number] = new_freq
10        self.count_map[old_freq] = self.count_map.get(old_freq, 0) - 1
11        if self.count_map[old_freq] == 0:
12            del self.count_map[old_freq]
13        self.count_map[new_freq] = self.count_map.get(new_freq, 0) + 1
14
15    def deleteOne(self, number):
16        old_freq = self.freq_map.get(number, 0)
17        if old_freq > 0:
18            new_freq = old_freq - 1
19            self.freq_map[number] = new_freq
20            self.count_map[old_freq] = self.count_map.get(old_freq, 0) - 1
21            if self.count_map[old_freq] == 0:
22                del self.count_map[old_freq]
23            if new_freq > 0:
24                self.count_map[new_freq] = self.count_map.get(new_freq, 0) + 1
25            else:
26                del self.freq_map[number]
27
28    def hasFrequency(self, frequency):
29        return frequency in self.count_map and self.count_map[frequency] > 0

ℹ

Complexity note: The time complexity is O(1) for each operation because we are using hash maps for constant time lookups and updates. The space complexity is O(n) due to storing frequencies and counts.

1Using hash maps allows for efficient frequency tracking.
2Maintaining a separate count of frequencies helps in quick queries.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.